258 lines
9.6 KiB
TeX
258 lines
9.6 KiB
TeX
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\documentclass[全部作业]{subfiles}
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\input{mysubpreamble}
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\begin{document}
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\setcounter{chapter}{6}
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\section{布朗运动及其分布}
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\subsection{练习题\thesection}
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（\textcolor{red}{若无特别说明}，以下总设$B=\{ B(t)\ ;\ t\geqslant 0 \}$为标准布朗运动。）
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\begin{enumerate}
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\questionandanswerSolution[2]{
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对任意$0<s<t$，求$B(s)+B(t)$的概率密度函数。
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}{
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首先，$B(s)$与$B(t)$不独立，所以需要进行变换，$B(s)+B(t)=2 B(s) + B(t)-B(s)$，而$B(s)$与$B(t)-B(s)$是独立的，且$B(s)\sim N(0,s)$, $B(t)-B(s)\sim N(0,t-s)$，所以
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$$
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B(s)+B(t)=2 B(s) + (B(t)-B(s)) \sim N(0, 4s+t-s)=N(0, 3s+t)
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$$
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所以$B(s)+B(t)$的概率密度函数为
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$$
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p(x) = \frac{1}{\sqrt{2\pi (3s+t)}} e^{- \frac{x^{2}}{3s+t}}
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$$
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}
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\questionandanswer[3]{
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在例6.1.9假设下，（例6.1.9：记$\mu=(1,2,3)^{\mathrm{T}}$, $\mathrm{D} = \begin{pmatrix}
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1 & 1 & 1 \\
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1 & 4 & 2 \\
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1 & 2 & 9 \\
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\end{pmatrix}$，$(X_1,X_2,X_3) \sim N(\mu, \mathrm{D})$）
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}{}
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\begin{enumerate}
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\questionandanswer[]{
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求$X_2=2$时$(X_1,X_3)$的条件联合分布；
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}{
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交换$X_2$与$X_3$的位置，$\displaystyle (X_1,X_3,X_2) \sim N\left( \begin{bmatrix}
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1 \\
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3 \\
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2 \\
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\end{bmatrix}, \begin{bmatrix}
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1 & 1 & 1 \\
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1 & 9 & 2 \\
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1 & 2 & 4 \\
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\end{bmatrix} \right) $，则$\tilde{\mu}=\begin{bmatrix}
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1 \\
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3 \\
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\end{bmatrix}, v=2$, $A=\begin{bmatrix}
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1 & 1 \\
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1 & 9 \\
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\end{bmatrix}, B = \begin{bmatrix}
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1 \\
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2 \\
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\end{bmatrix}, C=4$，
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$$
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\begin{aligned}
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(X_1,X_3)\sim &N(\tilde{\mu}+BC^{-1}(2-v), A-BC^{-1}B^{\mathrm{T}}) \\
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&=N\left( \begin{bmatrix}
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1 \\
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3 \\
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\end{bmatrix}, \begin{bmatrix}
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1 & 1 \\
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1 & 9 \\
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\end{bmatrix}-\begin{bmatrix}
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1 \\
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2 \\
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\end{bmatrix} \frac{1}{4} \begin{bmatrix}
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1 & 2 \\
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\end{bmatrix} \right) = N\left( \begin{bmatrix}
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1 \\
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3 \\
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\end{bmatrix}, \begin{bmatrix}0.75 & 0.5\\0.5 & 8.0\end{bmatrix} \right) \\
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\end{aligned}
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$$
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}
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\questionandanswer[]{
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求$X_2=2, X_3=3$时$X_1$的条件分布。
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}{
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$\tilde{\mu} = 1, \bar{y}=\begin{bmatrix}
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2 \\
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3 \\
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\end{bmatrix}, v=\begin{bmatrix}
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2 \\
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3 \\
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\end{bmatrix}$, $A=1, B=\begin{bmatrix}
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1 & 1 \\
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\end{bmatrix}, C=\begin{bmatrix}
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9 & 2 \\
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2 & 4 \\
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\end{bmatrix}$，
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$$
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\begin{aligned}
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X_1 \sim &N\left( \tilde{\mu}+BC^{-1}(\bar{y}-v), A-BC^{-1}B^{\mathrm{T}} \right) \\
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&=N\left( 1, 1-\begin{bmatrix}
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1 & 1 \\
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\end{bmatrix} \begin{bmatrix}
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9 & 2 \\
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2 & 4 \\
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\end{bmatrix}^{-1} \begin{bmatrix}
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1 \\
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1 \\
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\end{bmatrix} \right) =N\left( 1, \frac{23}{32} \right) \\
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\end{aligned}
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$$
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}
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\end{enumerate}
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\questionandanswer[4]{
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已知$B(1)=x$，对任意$0\leqslant s_1<s_2\leqslant s_3<s_4\leqslant 1$，证明
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}{}
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\begin{enumerate}
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\questionandanswerProof[]{
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$E((B(s_4)-B(s_3))(B(s_2)-B(s_1)))=(s_4-s_3)(s_2-s_1)(x^{2}-1)$,
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}{
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$$
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\begin{aligned}
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&E((B(s_4)-B(s_3))(B(s_2)-B(s_1))) \\
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&=\operatorname{Cov}(B(s_4)-B(s_3), B(s_2)-B(s_1)) + E(B(s_4)-B(s_3)) \cdot E(B(s_2)-B(s_1)) \\
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&=\operatorname{Cov}(B(s_4),B(s_2))-\operatorname{Cov}(B(s_4), B(s_1)) - \operatorname{Cov}(B(s_3), B(s_2))+\operatorname{Cov}(B(s_3), B(s_1)) \\ & + (E(B(s_4)) - E(B(s_3)))(E(B(s_2))-E(B(s_1))) \\
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\end{aligned}
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$$
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根据推论6.1.10，当$i<j$时，$\operatorname{Cov}(B(s_i),B(s_j))= s_i - s_i s_j$，$E(B(s_i))=s_i x$，所以
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$$
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\begin{aligned}
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&\text{上式}=s_2-s_2s_4 - s_1+s_1s_4-s_2+s_2s_3+s_1-s_1s_3 + (s_4 x-s_3x)(s_2x-s_1x) \\
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&= - s_{1} s_{3} + s_{1} s_{4} + s_{2} s_{3} - s_{2} s_{4} + x^{2} (s_{1} - s_{2}) (s_{3} - s_{4}) \\
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&=(s_4-s_3)(s_1-s_2)(x^{2}-1) \\
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\end{aligned}
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$$
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}
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\questionandanswerProof[]{
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$E\left[ (B(s_4)-B(s_1))^{2} \right] =(s_4-s_1)+(s_4-s_1)^{2}(x^{2}-1)$。
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}{
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仍然根据推论6.1.10，
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$$
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\begin{aligned}
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&E\left[ (B(s_4)-B(s_1))^{2} \right] =\operatorname{Var}[B(s_4)-B(s_1)]+[E(B(s_4)-B(s_1))]^{2} \\
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&=\operatorname{Var}[B(s_4)]+\operatorname{Var}[B(s_1)]-2\operatorname{Cov}(B(s_4),B(s_1))+[E(B(s_4))-E(B(s_1))]^{2} \\
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&=s_4-s_4^{2}+s_1-s_1^{2}-2(s_1+s_1s_4)+(s_4 x-s_1x)^{2} \\
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&=s_4-s_1+2s_1s_4-s_1^{2}-s_4^{2}+x^{2}(s_4-s_1) \\
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&=(s_4-s_1)+(s_4-s_1)(x^{2}-1) \\
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\end{aligned}
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$$
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}
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\end{enumerate}
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\questionandanswerSolution[5]{
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设$B$是方差参数$\sigma^{2}=3$的布朗运动，已知$B(1)=1,B(4)=4$，求$E(B(2)B(3))$和$E(B(3)B(5))$。
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}{
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由于$(B(1),B(2),B(3),B(4),B(5)) \sim N(\overrightarrow{0}, (i \land j)_{i,j=1,2,3,4})$，所以
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$$
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(B(2),B(3),B(5),B(1),B(4)) \sim N \left( \begin{bmatrix}
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0 \\
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0 \\
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0 \\
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0 \\
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0 \\
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\end{bmatrix}, 3\begin{bmatrix}
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2 & 2 & 2 & 1 & 2 \\
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2 & 3 & 3 & 1 & 3 \\
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2 & 3 & 5 & 1 & 4 \\
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1 & 1 & 1 & 1 & 1 \\
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2 & 3 & 4 & 1 & 4 \\
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\end{bmatrix} \right)
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$$
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由于$B(1)=1,B(4)=4$，所以
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$$
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\begin{aligned}
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(B(2),B(3))\sim &N (\tilde{\mu}+ BC^{-1}(\bar{y}-v), A-BC^{-1}B^{\mathrm{T}}) \\
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&=N\left( \begin{bmatrix}
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0 \\
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0 \\
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\end{bmatrix}+ \begin{bmatrix}
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1 & 2 \\
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1 & 3 \\
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\end{bmatrix} \begin{bmatrix}
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1 & 1 \\
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1 & 4 \\
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\end{bmatrix}^{-1} \left( \begin{bmatrix}
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1 \\
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4 \\
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\end{bmatrix}-\begin{bmatrix}
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0 \\
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0 \\
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\end{bmatrix} \right) , 3 \left(\begin{bmatrix}
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2 & 2 \\
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2 & 3 \\
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\end{bmatrix} - \begin{bmatrix}
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1 & 2 \\
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1 & 3 \\
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\end{bmatrix} \begin{bmatrix}
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1 & 1 \\
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1 & 4 \\
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\end{bmatrix}^{-1} \begin{bmatrix}
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1 & 2 \\
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1 & 3 \\
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\end{bmatrix} ^{\mathrm{T}} \right) \right) \\
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&=N\left( \begin{bmatrix}2\\3\end{bmatrix} , \begin{bmatrix}2 & 1\\1 & 2\end{bmatrix} \right) \\
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\end{aligned}
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$$
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所以
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$$
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\bm{E(B(2)B(3))} = \operatorname{Cov}(B(2),B(3))+E(B(2))E(B(3)) = 1 + 2 \times 3 \bm{=7}
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$$
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同理，
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$$
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\begin{aligned}
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(B(3),B(5)) \sim &N(\tilde{\mu}+BC^{-1}(\bar{y}-v), A-BC^{-1}B^{\mathrm{T}}) \\
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&=N\left( \begin{bmatrix}
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0 \\
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0 \\
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\end{bmatrix}+\begin{bmatrix}
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1 & 3 \\
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1 & 4 \\
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\end{bmatrix} \begin{bmatrix}
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1 & 1 \\
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1 & 4 \\
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\end{bmatrix}^{-1}\left( \begin{bmatrix}
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1 \\
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4 \\
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\end{bmatrix}-\begin{bmatrix}
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0 \\
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0 \\
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\end{bmatrix} \right) , 3 \left(\begin{bmatrix}
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3 & 3 \\
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3 & 5 \\
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\end{bmatrix}- \begin{bmatrix}
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1 & 3 \\
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1 & 4 \\
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\end{bmatrix} \begin{bmatrix}
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1 & 1 \\
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1 & 4 \\
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\end{bmatrix}^{-1} \begin{bmatrix}
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1 & 1 \\
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3 & 4 \\
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\end{bmatrix} \right) \right) \\
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&=N\left( \begin{bmatrix}3\\4\end{bmatrix} , \begin{bmatrix}2 & 0\\0 & 3\end{bmatrix} \right) \\
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\end{aligned}
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$$
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所以
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$$
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\bm{E(B(3)B(5))} = \operatorname{Cov}(B(3),B(5))+E(B(3))E(B(5) ) =0 + 3\times 4 \bm{= 12}
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$$
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}
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\questionandanswerProof[7]{
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令$Y(t)=B^{2}(t)-t$, $R(t)=e^{c B(t)-c^{2}t /2}$，其中$c >0$为常数。证明对任意$0\leqslant t<s$，$E(Y(s)|B(t))=Y(t)$ 以及 $E(R(s)|B(t))=R(t)$。
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}{
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根据推论6.1.10，由于$t<s$，所以$E(B(s)|B(t))=B(t), \operatorname{Var}(B(s)|B(t))=s-t$，所以
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$$
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\begin{aligned}
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&\bm{E(Y(s)|B(t))}=E(B^{2}(s)-s|B(t)) = E(B^{2}(s)|B(t)) - E(s|B(t)) \\
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=&\operatorname{Var}(B(s)|B(t))+\left( E(B(s)|B(t)) \right) ^{2} - s \\
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=&s-t+(B(t))^{2}-s = B^{2}(t) - t \bm{=T(t)} \\
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\end{aligned}
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$$
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$$
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\begin{aligned}
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&\bm{E(R(s)|B(t))} = E\left( e^{c B(s)-\frac{c^{2} s}{2}} \middle| B(t) \right) = E\left( e^{c B(s)}\middle| B(t) \right) E\left( e^{- \frac{c^{2}s}{2}} \middle| B(t) \right) \\
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=& E\left( e^{cB(t)}\ \middle|\ B(t) \right) E\left( e^{c(B(s) - B(t))} \middle| B(t) \right) E\left(e^{- \frac{c^{2}s}{2}}\ \middle|\ B(t)\right) \\
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=&e^{cB(t)} e^{\frac{c^{2}(s-t)}{2}} e^{-\frac{c^{2}s}{2}} =e^{cB(t) - \frac{c^{2} t}{2}} \bm{= R(t)} \\
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\end{aligned}
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$$
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}
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\end{enumerate}
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\end{document}
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